Problem: Simplify and expand the following expression: $ \dfrac{1}{2k + 18}- \dfrac{5}{3k - 24}- \dfrac{2}{k^2 + k - 72} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $2$ out of denominator in the first term: $ \dfrac{1}{2k + 18} = \dfrac{1}{2(k + 9)}$ We can factor a $3$ out of denominator in the second term: $ \dfrac{5}{3k - 24} = \dfrac{5}{3(k - 8)}$ We can factor the quadratic in the third term: $ \dfrac{2}{k^2 + k - 72} = \dfrac{2}{(k + 9)(k - 8)}$ Now we have: $ \dfrac{1}{2(k + 9)}- \dfrac{5}{3(k - 8)}- \dfrac{2}{(k + 9)(k - 8)} $ The least common multiple of the denominators is: $ 6(k + 9)(k - 8)$ In order to get the first term over $6(k + 9)(k - 8)$ , multiply by $\dfrac{3(k - 8)}{3(k - 8)}$ $ \dfrac{1}{2(k + 9)} \times \dfrac{3(k - 8)}{3(k - 8)} = \dfrac{3(k - 8)}{6(k + 9)(k - 8)} $ In order to get the second term over $6(k + 9)(k - 8)$ , multiply by $\dfrac{2(k + 9)}{2(k + 9)}$ $ \dfrac{5}{3(k - 8)} \times \dfrac{2(k + 9)}{2(k + 9)} = \dfrac{10(k + 9)}{6(k + 9)(k - 8)} $ In order to get the third term over $6(k + 9)(k - 8)$ , multiply by $\dfrac{6}{6}$ $ \dfrac{2}{(k + 9)(k - 8)} \times \dfrac{6}{6} = \dfrac{12}{6(k + 9)(k - 8)} $ Now we have: $ \dfrac{3(k - 8)}{6(k + 9)(k - 8)} - \dfrac{10(k + 9)}{6(k + 9)(k - 8)} - \dfrac{12}{6(k + 9)(k - 8)} $ $ = \dfrac{ 3(k - 8) - 10(k + 9) - 12} {6(k + 9)(k - 8)} $ Expand: $ = \dfrac{3k - 24 - 10k - 90 - 12}{6k^2 + 6k - 432} $ $ = \dfrac{-7k - 126}{6k^2 + 6k - 432}$